Question

Solve the quadratic polynomial \[4{x^2} + 20x + 25\].

Answer 1

Hint: To solve $a{x^2} + bx + c$, we have to write it in a simplified manner. It can be done by factoring the terms and if the factored form has the same factors, then it can be written in the whole square form.
We will use the factorization method to solve it.
For this we have to write $a{x^2} + bx + c$ in the form of $(x - \alpha )(x - \beta )$ where $\alpha $ and $\beta $ are evaluated using by splitting the term $bx$ in the form of ${b_1}x + {b_2}x$ such that the sum of ${b_1}$ and ${b_2}$ is $b$ and the product of ${b_1}$ and ${b_2}$ is $a \times c$.
Then further simplifying the terms we get $(x - \alpha )(x - \beta )$.

Complete step-by-step solution:
We have the following term:
  \[4{x^2} + 20x + 25\]
To solve the problem, compare it with the standard form $a{x^2} + bx + c$,
So, we get
$a = 4,b = 20,c = 25$
Now, we need to split the term $bx$ in ${b_1}x + {b_2}x$ such that the sum of ${b_1}$ and ${b_2}$ is $b$ and the product of ${b_1}$ and ${b_2}$ is $a \times c$.
Using the given terms, we get
\[{b_1} + {b_2} = 20\]
\[{b_1} \times {b_2} = 4 \times 25 = 100\]
Let us take \[{b_1} = 10\] and \[{b_2} = 10\], since it satisfies both the above conditions,
\[{b_1} + {b_2} = 10 + 10 = 20\]
\[{b_1} \times {b_2} = 10 \times 10 = 100\]
We split $20x$ term into $10x$ and $10x$,
\[4{x^2} + 10x + 10x + 25\]
Take the term $2x$ common from the terms $4{x^2} + 10x$ and take $5$ common from the terms $10x + 25$, to obtain
\[2x(2x + 5) + 5(2x + 5)\]
Take the term $2x + 5$common from the above terms, we get
$(2x + 5)(2x + 5)$
Since, both the terms are same so it can be written in the form of square, so we get
${(2x + 5)^2}$
So, the term \[4{x^2} + 20x + 25\] can be written in the factored form as ${(2x + 5)^2}$.

Note: The problem $a{x^2} + bx + c$ has to be written in the factored form $(x - \alpha )(x - \beta )$. It has to be kept in mind that the term $bx$ is split into ${b_1}x + {b_2}x$ such that the sum of ${b_1}$ and ${b_2}$ is $b$ and the product of ${b_1}$ and ${b_2}$ is $a \times c$. Then further simplifying gives the factored form.